So often, the lonely single post! Well, here I am now to keep you company.

I ALWAYS solve a puzzle before looking at the discussion. But then I'm always glad for confirmation that I got it right or (sometimes) that I had missed something easier.

Thank you! Keep it up.

One day I'll be the first to post. I did that just once before, taking advantage of my time zone. To be first to post I'd have to get to work straight away after (my) midnight, and that doesn't happen often!

There is also an interesting chain that forces r3c8 to be one, both when r1c7=4 and also ...
If r1c7=6 then r8c6=4 so r9c8=6 and r9c1=5 so r9c2=4 and r7c2=3
so r4c2=9 and r4c4=8 which makes r3c4=4 SO r3c8 <>4
AND the same logic also makes r4c6=3 and r2c6=4 which would mean that there were two occurrences of 4 in Box 2 which is not allowed so r1c7<>6

This is some kind of a chain: does it have a technical name, or is it more in the realm of a guess?

There is also an interesting chain that forces r3c8 to be one, both when r1c7=4 and also ...
If r1c7=6 then r8c6=4 so r9c8=6 and r9c1=5 so r9c2=4 and r7c2=3
so r4c2=9 and r4c4=8 which makes r3c4=4 SO r3c8 <>4
AND the same logic also makes r4c6=3 and r2c6=4 which would mean that there were two occurrences of 4 in Box 2 which is not allowed so r1c7<>6

This is some kind of a chain: does it have a technical name, or is it more in the realm of a guess?

Terence,

I'm assuming the 2nd term should be r8c7 rather than r8c6. Then you have an XY-Chain, a series of bivalue cells where you start with a cell XY where the X proves a Y in another cell. Any cell seeing both the start and end cells of the chain cannot be Y, thus r3c8<>4.

After that I don't understand. Once the 4 is gone from r3c8 the 4's in r3 are confined to box 2, thus no other lines in box 2 can be 4, so r2c6<>4=3.

Terence, here is the Eureka notation of your "XY-Chains" :
(1=4)r3c8-(4=6)r1c7-(6=4)r8c7-(4=6)r9c8-(6=5)r9c1-(5=4)r9c2-(4=3)r7c2-(3=9)r4c2-(9=8)r4c4-[(8=4*)r3c4-(4=1)r3c8 and (8=3)r4c6-(3=4*)r2c6]
If you are not used to such a notation, the following indications should be sufficient to understand it :

a. The reading is from left to right;
b. a=A : a False => A True;
c. a-A : a True => A False.

Now,

1. in the first "and" case, r3c8<>1 => r3c8=1;
2. the contradiction observed in the 2 "and" cases => r3c8<>1 is false;
3. "the XY Wing" is contained in the 2 "and" cases inside the brakets : r3c4=8 and r4c6=8 cannot be false together => r4c4<>8.

As a more general way to see what you have so interestingly found, just consider that the cells in R469+C47+B49+[B7-NT(129)] have only 2 solutions.
They are identified by the letters a and A in the following matrix :